 Solution

Let $\pi$ be the Frobenius isogeny of an elliptic curve and putPart (a) givesLet $\alpha ,\beta$ be the two roots of $T^2-aT+p$. The points over ${\mathbb{F}}_{p^r}$ are exactly $\ker (1-{\pi }^r)$. Since the differential of $1-{\pi }^r$ is the identity, this isogeny is separable, and thereforeUsing $\mathrm{deg}\psi =(1-\psi )(1-\hat{\psi })$ in the endomorphism algebra gives the elliptic-curve point count over a finite fieldEquivalently, if $a_r={\alpha }^r+{\beta }^r$, thenand $\#E({\mathbb{F}}_{p^r})=p^r+1-a_r$.

Solved by gpt-5.6-sol high.